Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{4k^2 - 20k - 144}{k^2 - 11k + 18} \div \dfrac{k + 4}{k + 1} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{4k^2 - 20k - 144}{k^2 - 11k + 18} \times \dfrac{k + 1}{k + 4} $ First factor out any common factors. $n = \dfrac{4(k^2 - 5k - 36)}{k^2 - 11k + 18} \times \dfrac{k + 1}{k + 4} $ Then factor the quadratic expressions. $n = \dfrac {4(k - 9)(k + 4)} {(k - 9)(k - 2)} \times \dfrac {k + 1} {k + 4} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac { 4(k - 9)(k + 4) \times (k + 1)} { (k - 9)(k - 2) \times (k + 4)} $ $n = \dfrac {4(k - 9)(k + 4)(k + 1)} {(k - 9)(k - 2)(k + 4)} $ Notice that $(k - 9)$ and $(k + 4)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {4\cancel{(k - 9)}(k + 4)(k + 1)} {\cancel{(k - 9)}(k - 2)(k + 4)} $ We are dividing by $k - 9$ , so $k - 9 \neq 0$ Therefore, $k \neq 9$ $n = \dfrac {4\cancel{(k - 9)}\cancel{(k + 4)}(k + 1)} {\cancel{(k - 9)}(k - 2)\cancel{(k + 4)}} $ We are dividing by $k + 4$ , so $k + 4 \neq 0$ Therefore, $k \neq -4$ $n = \dfrac {4(k + 1)} {k - 2} $ $ n = \dfrac{4(k + 1)}{k - 2}; k \neq 9; k \neq -4 $